Multiple choice

If both the mean and the standard deviation of $50$ observations ${ x }{ 1 },{ x }{ 2 },......,{ x }{ 50 }$ are equal to $16$, then the mean of ${ \left( { x }{ 1 }-4 \right) }^{ 2 },{ \left( { x }{ 2 }-4 \right) }^{ 2 },.....{ \left( { x }{ 50 }-4 \right) }^{ 2 }$ is:

  1. $525$
  2. $380$
  3. $480$
  4. $400$
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D Correct answer
Explanation

The mean of (xi - 4)^2 is the variance of the original set plus the square of the difference between the mean and 4. Specifically, E[(X-4)^2] = E[X^2 - 8X + 16] = E[X^2] - 8E[X] + 16. Since Variance = E[X^2] - (E[X])^2, we have E[X^2] = 16^2 + 16^2 = 512. Substituting: 512 - 8(16) + 16 = 512 - 128 + 16 = 400.

AI explanation

Using the formula for the mean of squared deviations, the average of (x minus 4) squared terms expands to the mean of x squared minus 8 times the mean of x plus 16. The variance is given as 16 squared, which is 256, and since the mean is 16, the mean of x squared equals 256 plus 256, resulting in 512. Substituting these values gives 512 minus 8 times 16 plus 16, which equals 512 minus 128 plus 16, resulting in a mean of 400.

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